Date: 02/27/2011 at 23:28:24
From: Doctor Vogler
Subject: Re: how to find integer number accomplish (x^2+y)(x+y^2)=(x-y)3
Hi,
Thanks for writing to Dr. Math. That's an interesting curve.
You might be interested in its parameterization, or the way to find all
rational-number solutions. It turns out that (-1, -1) is a special
solution to this equation.
Suppose you have a rational-number solution (x, y) = (r, s). Then consider
the line ...
y = mx + b
... that goes through both (-1, -1) and (r, s). In other words, you take
m = (s + 1)/(r + 1)
b = (s - r)/(r + 1).
In fact, let's consider all lines with rational slope that pass through
(-1, -1). That is, all lines of the form
y = m(x + 1) - 1.
Now let's look at the points that intersect both this line and the curve
(x^2 + y)(x + y^2) = (x - y)^3.
Substituting m(x + 1) - 1 for y in the equation for the curve, we get
(x^2 + m(x + 1) - 1)(x + (mx + m - 1)^2) = (x - (mx + m - 1))^3.
We know that x = -1 is going to be a solution of the above equation. It
turns out that it's a double root. And y = 0 is also a solution. So we can
simplify the above equation into
(x + 1)^2*(m(x + 1) - 1)*(mx + 2(m - 1)^2).
So apart from the known solution (-1, -1), the other two solutions to both
the line and the curve are
(1/m - 1, 0)
(-2(m - 1)^2/m, (m - 1)(3 - 2m)).
So it turns out that your curve is really a line and a twisted curve
(which has two pieces, one for m positive and one for m negative). The
line is y = 0. The twisted curve is parameterized by ...
x = -2(m - 1)^2/m
y = (m - 1)(3 - 2m),
... which crosses over itself at m = 1/2 and m = 2, where
(x, y) = (-1, -1). By letting m run over all real numbers, you trace out
the curve. By letting m be any rational number, you get all rational
solutions to your equation.
But you want integer solutions. Well, integer numbers are rational
numbers, too. So if (x, y) is an integer-number solution, then either
y = 0, or ...
x = -2(m - 1)^2/m
y = (m - 1)(3 - 2m)
... for some rational number m. Write m in reduced form as m = r/s, where
r and s are integers, s > 0, and r and s have no common factors larger
than 1. Then it follows that
x = -2(r - s)^2/(rs)
y = (r - s)(3s - 2r)/s^2.
But (r - s)^2/(rs) is in reduced form, because if there were any prime
number p that was a factor of both the denominator, rs, and the numerator,
(r - s)^2, then p would have to be a factor of either r or s (in order to
divide the denominator) and also of their difference, r - s (in order to
divide the numerator). Therefore, it would have to be a factor of both r
and s, contradicting that m = r/s is in reduced form.
So if x is to be an integer, then rs must be a factor of 2, and there are
only four integers that divide 2 -- namely, +2, +1, -1, and -2. So the
only six possibilities are
(r, s) = (1, 1)
(r, s) = (-1, 1)
(r, s) = (2, 1)
(r, s) = (-2, 1)
(r, s) = (1, 2)
(r, s) = (-1, 2).
And that gives you all integer-number solutions to your equation, apart
from the y = 0 solutions -- namely,
(x, y) = (0, 0)
(x, y) = (8, -10)
(x, y) = (-1, -1)
(x, y) = (9, -21)
(x, y) = (-1, -1)
(x, y) = (9, -6)
Of course, one of those is also a y = 0 solution, and two are duplicates
(the curve crosses over itself), so the integer solutions are:
(x, y) = (-1, -1)
(x, y) = (8, -10)
(x, y) = (9, -21)
(x, y) = (9, -6)
(x, y) = (anything, 0)
If you have any questions about this or need more help, please write back
and show me what you have been able to do, and I will try to offer further
suggestions.
- Doctor Vogler, The Math Forum
http://mathforum.org/dr.math/

Date: 02/28/2011 at 04:05:34
From: swidan
Subject: Thank you (how to find integer number accomplish (x^2+y)(x+y^2)=(x-
y)3)
Thank you so much for the help.